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Saturday, August 18, 2007

M Theory Lesson 87

In a 2002 paper, Kennison proves the following result. First, let Stone be the category of Stone spaces and Bool be the category of Boolean algebras. There is a categorical equivalence $P$ from Bool to Stone which takes the space of all points in an object $B$. An arrow $t: X \rightarrow X$ in Stone is called Boolean cyclic if the corresponding arrow in Bool is cyclic, which means that the supremum over all equalisers of $t^{n}$ and $1_{X}$ is just $X$. Intuitively, these represent dynamical processes that eventually cycle. Now let $\mathbb{N}$ be the ordinals. An action is a map $\mathbb{N} \times X \rightarrow X$ given by $(n,x) \mapsto (t^{n}(x))$. Kennison showed that the property of being Boolean cyclic was equivalent to actually having an action by Z, the profinite integers, namely the product over ordinal primes $\prod \mathbb{Z}_{p}$ of the p-adic integers.

This is interesting because J. Borger has been looking at finite sets with Z actions in order to characterise $\Lambda$-rings in terms of finite sets with actions of $G \times \mathbb{N}^{+}$ for $G$ the absolute Galois group for the rationals. Recall that this group acts on Grothendieck's ribbon graphs. A $\Lambda$-ring structure is a series of arrows $f_{p}: R \rightarrow R$ for a ring $R$. For example, one may take

$R = \frac{\mathbb{Z} [x]}{x^r - 1}$

along with the Frobenius maps $\psi_{p}: x \mapsto x^p$. Borger et al show that certain nice $\Lambda$-rings can only be a field if they are in fact the rationals. It turns out that all nice $\Lambda$-rings are subrings of products of the cyclotomic example above, for some $r$. This seems to be important somehow...

Hi Doug. It is certainly true, and a great tragedy, that in my generation physicists got away with almost no education in mathematics. I have had to teach myself almost everything I know (which is therefore frightfully little - sigh). Fortunately, the modern syllabus is adapting slowly to the times.